AMC10 2020 A

Adding $\frac{3}{4}$ to both sides of the equation gives $x = \frac{5}{12} - \frac{1}{3} + \frac{3}{4} = \frac{5}{12} - \frac{4}{12} + \frac{9}{12} = \frac{10}{12} = \frac{5}{6}$.

Because the average of the five numbers is 15, $3 + 5 + 7 + a + b = 5 \cdot 15 = 75$, so $a + b = 75 - (3 + 5 + 7) = 60$. Therefore the average of $a$ and $b$ is $\frac{1}{2}(a+b) = 30$.

Because $\frac{x-y}{y-x} = \frac{x-y}{-(x-y)} = -1$, the expression simplifies to $\frac{a-3}{5-a} \cdot \frac{b-4}{4-b} \cdot \frac{c-5}{5-c} = (-1)^3 = -1$.

The driver travels $2 \cdot 60 = 120$ miles, which requires $120 \div 30 = 4$ gallons of gasoline. Her gasoline expense is $4 \cdot 2 = 8$. She is paid $120 \cdot 0.50 = 60$, so after her gasoline expense, she makes $60 - 8 = 52$ in 2 hours, which is $26$ per hour.

The given equation is equivalent to $x^2 - 12x + 34 = \pm 2$, which splits into two cases, $x^2 - 12x + 32 = 0$ and $x^2 - 12x + 36 = 0$. Because $x^2 - 12x + 32 = (x-4)(x-8)$, values of $x$ that satisfy the first equation are 4 and 8; and because $x^2 - 12x + 36 = (x-6)^2$, the only solution of the second equation is 6. The sum of these three values is $4 + 8 + 6 = 18$.

For an integer to satisfy the given conditions, the thousands digit must lie in $\lbrace 2, 4, 6, 8\rbrace$, the hundreds and tens digits must lie in $\lbrace 0, 2, 4, 6, 8\rbrace$, and the units digit must be 0. By the Multiplication Principle for counting, the number of choices of digits is $4 \cdot 5 \cdot 5 \cdot 1 = 100$.

Answer (C): The sum of all 25 integers from $-10$ to $14$, inclusive, equals $11+12+13+14=50$ because the integers in that sum from $1$ to $10$ can be paired with their additive inverses to contribute $0$ to the sum. Because each of the $5$ rows in the $5$-by-$5$ square has the same sum and there are $5$ rows, that common sum is $50 \div 5 = 10$. Here is one such magic square, in which each row, each column, and each main diagonal sums to $10$. \[ \begin{array}{|c|c|c|c|c|} \hline 6 & 13 & -10 & -3 & 4\\ \hline 12 & -6 & -4 & 3 & 5\\ \hline -7 & -5 & 2 & 9 & 11\\ \hline -1 & 1 & 8 & 10 & -8\\ \hline 0 & 7 & 14 & -9 & -2\\ \hline \end{array} \]

Answer (B): When terms are combined in groups of four, the sum is the arithmetic series $2+10+18+\cdots+394$ with 50 terms. Its sum is $\dfrac{50}{2}\cdot(2+394)=9,900.$ OR The sum can be viewed as $\displaystyle \sum_{k=1}^{200}k-2\sum_{k=1}^{50}4k=\dfrac{200\cdot201}{2}-8\cdot\dfrac{50\cdot51}{2}=9,900.$

One adult and one child will occupy $\frac{1}{7} + \frac{1}{11} = \frac{18}{77}$ of a bench section. The problem is asking for the least possible positive integer value of $N$ that will make $\frac{77}{18} \cdot N$ an integer. Because 77 and 18 are relatively prime, the least such value is $N = 18$, in which case 77 adults and 77 children will occupy all the bench space.

Answer (B): Because the volumes of the cubes are given to be $1^3, 2^3, \ldots, 7^3$, the edge lengths are $1, 2, \ldots, 7$. If the cubes were not stacked, the total surface area would be $6\cdot(1^2+2^2+\cdots+7^2)$. When two cubes meet along a surface, two of the smaller face areas are lost, for a total area of $2\cdot(1^2+2^2+\cdots+6^2)$ that needs to be subtracted from the surface area of the unstacked cubes to get the total surface area of the tower. The result is $$ 4\cdot(1^2+2^2+\cdots+6^2)+6\cdot 7^2 =4\cdot 91+6\cdot 49 =658. $$

Because there are 4040 numbers in the list, the median is the mean of the numbers in positions 2020 and 2021 when the list is put into nondecreasing order. All of the squared entries are greater than all of the non-squared entries except for $1^2, 2^2, \dots, 44^2 = 1936$. When the numbers are put into nondecreasing order, those 44 entries will be in the first half, and the greatest 44 non-squared entries, namely 1977, 1978, \dots, 2020, will be in the second half. Therefore the entry in position 2020 will be 1976, the entry in position 2021 will be 1977, and the median will be 1976.5.

Let $G$ be the point of intersection of $\overline{MV}$ and $\overline{CU}$. Right triangle $CVG$ has legs with lengths $GC = \frac{3}{2} \cdot 12 = 8$ and $GV = \frac{1}{2} \cdot 12 = 4$, so it has area $\frac{1}{2} \cdot 8 \cdot 4 = 16$. Because the medians of a triangle divide the triangle into six triangular regions of equal area, the area of $\triangle AMC$ is $6 \cdot 16 = 96$.

In the figure below, the solid circle is the starting point and the open circles are the possible ending points. After the first jump, the frog is at one of the points (0, 2), (1, 1), (1, 3), or (2, 2), each with equal probability. If it is at (0, 2), then the sequence of jumps ends, and the frog is on a vertical side of the square. In the other three cases, the frog is on a diagonal of the square, so by symmetry the sequence is equally likely to end on a horizontal side or a vertical side of the square. Thus the requested probability is $\frac{1}{4} \cdot 1 + \frac{3}{4} \cdot \frac{1}{2} = \frac{5}{8}$.

Answer (D): The given expression equals $$ \frac{x^3y^2+x^5+y^5+y^3x^2}{x^2y^2}=\frac{(x^2+y^2)(x^3+y^3)}{(xy)^2}. $$ Note that $$ x^2+y^2=(x+y)^2-2xy=4^2-2(-2)=20. $$ Because $(x+y)^3=x^3+y^3+3x^2y+3xy^2$, it follows that $$ \begin{aligned} x^3+y^3&=(x+y)^3-3x^2y-3xy^2\\ &=(x+y)^3-3xy(x+y)\\ &=4^3-3(-2)(4)\\ &=88. \end{aligned} $$ The requested value is $$ \frac{20\cdot 88}{(-2)^2}=440. $$

Answer (E): The prime factorization of 12! = 12 · 11 · · · 1 is $2^{10}\cdot 3^{5}\cdot 5^{2}\cdot 7\cdot 11$. To choose one divisor at random is equivalent to choosing an exponent for 2 at random from $\{0,1,2,\ldots,10\}$, an exponent for 3 at random from $\{0,1,2,\ldots,5\}$, an exponent for 5 at random from $\{0,1,2\}$, and exponents for 7 and 11 at random from $\{0,1\}$. It follows that there are $(10+1)\cdot(5+1)\cdot(2+1)\cdot(1+1)\cdot(1+1)=11\cdot 6\cdot 3\cdot 2\cdot 2$ possible divisors. The chosen divisor will be a perfect square if and only if all the chosen exponents are even. There are 6 even choices for the exponent of 2 (namely 0, 2, 4, 6, 8, or 10), 3 even choices for the exponent of 3, 2 even choices for the exponent of 5, and 1 even choice for the exponent of 7 and the exponent of 11. The probability that the chosen divisor is a perfect square is therefore $\dfrac{6\cdot 3\cdot 2}{11\cdot 6\cdot 3\cdot 2\cdot 2}=\dfrac{1}{22}$. The requested sum is $1+22=23$.

Answer (B): Without loss of generality, it can be assumed that the point falls within the unit square with vertices $(0,0)$, $(1,0)$, $(1,1)$, and $(0,1)$. The probability that the point is within $d$ units of a lattice point is the total area of the four quarter-circles of radius $d$ centered at the corners of the square, provided $d<\frac12$. This is the shaded region in the figure. Thus $\pi d^2=\frac12$, so $d=\frac{1}{\sqrt{2\pi}}<\frac12$. Estimating gives $$ d=\frac{1}{\sqrt{2\pi}}\approx\frac{1}{\sqrt{6.28}}\approx\frac{10}{\sqrt{625}}=\frac{10}{25}=0.4. $$ For a rigorous justification of this estimate, it must be shown that $\frac{1}{\sqrt{2\pi}}$ satisfies $$ \frac{7}{20}=0.35<\frac{1}{\sqrt{2\pi}}<0.45=\frac{9}{20}, $$ which is equivalent to $$ \frac{49\pi}{400}<\frac12<\frac{81\pi}{400}. $$ Indeed, $$ \frac{49\pi}{400}<\frac{50\pi}{400}=\frac{\pi}{8}<\frac12 $$ and $$ \frac{81\pi}{400}>\frac{80\pi}{400}=\frac{\pi}{5}>\frac12. $$

Answer (E): Observe that $P(n)\le 0$ exactly when either one of the factors in the product is zero or an odd number of the factors are negative. Therefore $P(n)\le 0$ exactly when $n$ lies in one of the intervals $[1^2,2^2]$, $[3^2,4^2]$, $\ldots$, $[99^2,100^2]$. The number of integers in the interval $[(2k-1)^2,(2k)^2]$ is $(2k)^2-(2k-1)^2+1=4k$, so the total number of values for $n$ is \[ \sum_{k=1}^{50}4k=4\cdot\frac{50\cdot 51}{2}=5100. \]

Answer (C): A product of two integers is odd if and only if both factors are odd. Therefore there are $2\cdot 2=4$ ordered pairs $(a,d)$ such that $a\cdot d$ is odd. For the remaining $16-4=12$ ordered pairs, $a\cdot d$ is even. The same calculations apply to $b\cdot c$. The difference is odd if and only if one of the terms is odd and the other is even. Therefore there are $4\cdot 12+12\cdot 4=96$ ordered quadruples that make $a\cdot d-b\cdot c$ odd.

Answer (E): Shown below is a planar graph representation of the dodecahedron with the top face in the center and the edges of the bottom face forming the border of the figure. Starting from the top face (the inner pentagon in this figure), there are 5 possible moves down to the next level (the top ring, which consists of the unshaded pentagons in the figure). For the next few moves there are $1+2\cdot 4=9$ possibilities because one can move directly down to the next lower level (the bottom ring), or one can remain on the same level and make 1 to 4 moves in either direction around the top ring before moving down. When the path moves to the bottom ring, there are 2 possible faces to choose from. Then again there are 9 possibilities for how the path moves before reaching the bottom face. This gives a total of $5\cdot 9\cdot 2\cdot 9=810$ possible paths from the top face to the bottom face.

Answer (D): Because the diagonals intersect, quadrilateral $ABCD$ is convex. Let $F$ be the foot of the perpendicular from $B$ to $AC$. Because $\triangle ECD$ is a right triangle, $\angle AEB=\angle DEC$ is acute, so $F$ lies between $A$ and $E$. Let $h=BF$ and $x=EF$. See the figure below. Because $\triangle BFE\sim\triangle DCE$, $$ \frac{h}{x}=\frac{30}{20-5}=2. $$ Because $\triangle ABF\sim\triangle BCF$, $$ \frac{h}{5-x}=\frac{15+x}{h}. $$ Together these imply that $x^2+2x-15=0$. Solving for $x$ yields $x=3$ and $h=6$. The area of $ABCD$ is the area of $\triangle ABC$ plus the area of $\triangle ACD$, and this equals $\frac12\cdot 20\cdot 6+\frac12\cdot 20\cdot 30=360$.

Answer (C): Note that the problem is equivalent to finding the number of nonzero digits when the left-hand side is written in binary. More generally, consider the fraction $$ \frac{2^{n^2}+1}{2^n+1} $$ for odd $n$. By the formula for the sum of a geometric series, $$ \frac{2^{n^2}+1}{2^n+1}=2^{(n-1)n}-2^{(n-2)n}+2^{(n-3)n}-\cdots+2^{2n}-2^n+1. $$ This expression is not in the form of the problem statement because of the negative signs, but it is possible to transform it into such a form by considering pairs of adjacent terms. For $i=1,3,5,\ldots,n-2$, $$ 2^{(i+1)n}-2^{in}=2^{in}(2^n-1) =2^{in}(1+2+\cdots+2^{n-1}) =2^{in}+2^{in+1}+\cdots+2^{(i+1)n-1}. $$ Thus each pair $2^{(i+1)n}-2^{in}$ contributes $((i+1)n-1)-in+1=n$ nonzero digits to the binary expansion of the left-hand side. The total number of expressions of this form is equal to the number of integers in the set $\{1,3,5,\ldots,n-2\}$, which is $\frac12(n-1)$. Thus, including the last term of $1$, the total number of summands is $n\cdot\frac12(n-1)+1$. For $n=17$, this value is $17\cdot 8+1=137$.

Answer (A): If $n$ is a divisor of $999$, then $\left\lfloor \frac{999}{n}\right\rfloor = 1 + \left\lfloor \frac{998}{n}\right\rfloor$. Otherwise, $\left\lfloor \frac{999}{n}\right\rfloor = \left\lfloor \frac{998}{n}\right\rfloor$. Similarly, $\left\lfloor \frac{1000}{n}\right\rfloor = 1 + \left\lfloor \frac{999}{n}\right\rfloor$ if $n$ is a divisor of $1000$, and $\left\lfloor \frac{1000}{n}\right\rfloor = \left\lfloor \frac{999}{n}\right\rfloor$ otherwise. Therefore if $n$ divides neither $999$ nor $1000$, then the given sum has the form $a+a+a=3a$, so it is a multiple of $3$. Further, if $n$ divides both $999$ and $1000$ (that is, if $n=1$), then the sum has the form $a+(a+1)+(a+2)=3a+3$, which is also a multiple of $3$. Thus in order for the sum not to be a multiple of $3$, $n$ must be a divisor of either $999$ or $1000$, but not both, so that the sum has the form $a+a+(a+1)=3a+1$ or $a+(a+1)+(a+1)=3a+2$. Because $999=3^3\cdot 37^1$ and $1000=2^3\cdot 5^3$, it follows that $999$ has $(3+1)(1+1)=8$ divisors and $1000$ has $(3+1)(3+1)=16$ divisors. As $n\ne 1$, the number of possible values for $n$ is $(8-1)+(16-1)=22$.

Answer (A): If two rotations that are not inverses of each other are applied in order, then there is a unique third rotation that will return $T$ to its original position. There are $3\cdot2=6$ of these. If two different reflections across the coordinate axes are applied in order, then the $180^\circ$ rotation is the unique transformation that will return $T$ to its original position. There are $2$ of these. If the first two transformations are a rotation of $180^\circ$ and a reflection, in either order, then the other reflection is the unique transformation that will return $T$ to its original position. There are $4$ such sequences. It remains to show that for all other choices of the first two transformations, none of the five choices for the third transformation will return $T$ to its original position. Two transformations that are inverses of each other already return $T$ to its original position, so applying a third transformation will move $T$ out of its original position. The other cases are a rotation of $90^\circ$ or $270^\circ$ followed by a horizontal or vertical reflection, or vice versa. In each case, a reflection would be needed to restore the correct orientation of the triangle, but neither a vertical reflection nor a horizontal reflection will return $T$ to its original position. In all, there are $6+2+4=12$ sequences of three of the given transformations that will return $T$ to its original position.

Answer (C): The condition $\gcd(n+63,120)=60$ is true if and only if $n+63$ is divisible by $60$ but not $120$, which is equivalent to $n \equiv 57 \pmod{60}$ and $n \equiv 117 \pmod{120}.$ Similarly, the condition $\gcd(63,n+120)=21$ is true if and only if $n+120$ is divisible by $21$ but not $63$, which is equivalent to $n \equiv 6 \pmod{21}$ and $n \equiv 27 \text{ or } 48 \pmod{63}.$ From the congruences $n \equiv 117 \pmod{120}$ and $n \equiv 6 \pmod{21}$ it follows that $n \equiv 237 \pmod{840}$. Thus $n=237+840k$ for some integer $k$, and because $n$ was specified to be greater than $1000$, $k$ must be at least $1$. The condition $n \equiv 57 \pmod{60}$ holds for all such $n$. However, the condition $n \equiv 27$ or $48 \pmod{63}$ does not hold when $k=1$ and $n=1077$, but it does hold when $k=2$ and $n=1917$. The requested sum of digits of $n$ is $1+9+1+7=18$.

Answer (A): There are 15 ways to roll a 7 with three dice (5-1-1, 4-2-1, 3-3-1, and 3-2-2 and their permutations) out of the $6^3$ possible rolls, so the probability of rolling 7 with three dice is $p_3=\frac{5}{72}$. Also, for $1\le a\le 7$, the probability of rolling $a$ with two dice is $p_2(a)=\frac{a-1}{36}$, and the probability of rolling any number on a single die is $p_1=\frac{1}{6}$. Because $p_2(a)<p_1$ for all $a<7$, Jason will always choose to reroll one die instead of two if possible; that is, if any two of the initial rolls sum to at most 6, then Jason is better off keeping those two rolls and rerolling the third, rather than keeping just one roll. Therefore Jason will choose to reroll two of the dice only if every pair of the initial rolls sums to at least 7. Furthermore, because $p_2(2)<p_3$ and $p_2(3)<p_3$, if all the rolls are at least 4, then Jason will choose to reroll all three dice. Therefore Jason will reroll two dice if and only if each pair of the initial rolls sums to at least 7, but at least one of the rolls is at most 3. These possibilities are given by 1-6-6, 2-$a$-$b$ where $a,b\in\{5,6\}$, and 3-$c$-$d$ where $c,d\in\{4,5,6\}$. Including permutations, this gives $3+12+27=42$ possibilities, so the requested probability is $\frac{42}{216}=\frac{7}{36}$.